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A gas expands against a non-zero external pressure while in thermal isolation from the surroundings. For this expansion a. \(\Delta \mathrm{E}=\mathrm{q}\) b. \(\Delta \mathrm{E}\) increases c. \(\Delta \mathrm{E}\) does not change d. \(\Delta \mathrm{E}\) decreases

Short Answer

Expert verified
\( \Delta \mathrm{E} \) decreases (option d).

Step by step solution

01

Understanding Key Concepts

This question involves thermodynamic principles, specifically the First Law of Thermodynamics, which is stated as \( \Delta \mathrm{E} = \mathrm{q} + \mathrm{w} \). Here, \( \Delta \mathrm{E} \) is the change in internal energy of the system, \( \mathrm{q} \) represents heat exchange, and \( \mathrm{w} \) represents work done. If the system is thermally isolated, \( \mathrm{q} = 0 \).
02

Evaluate Heat Exchange

Since the system is in thermal isolation, no heat is exchanged with the surroundings. This means \( \mathrm{q} = 0 \).
03

Analyze Work Done by the System

When a gas expands against an external pressure, it does work on the surroundings; this work is generally negative from the system's perspective because energy is leaving the system. Therefore, the expression becomes \( \Delta \mathrm{E} = 0 + \mathrm{w} \) where \( \mathrm{w} \) is negative.
04

Assess the Change in Internal Energy

The internal energy change \( \Delta \mathrm{E} \) is determined by the work done. Since \( \mathrm{w} \) is negative, \( \Delta \mathrm{E} \) is also negative, indicating a decrease in internal energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Internal Energy Change
The concept of internal energy change is fundamental to understanding many processes in thermodynamics. Internal energy, denoted as \( \Delta \mathrm{E} \), is the total energy contained within a system, accounting for both kinetic and potential energy of particles. In a chemical or physical process, the change in this internal energy can signify energy being added to or removed from the system, commonly involving heat exchange or work done.

According to the First Law of Thermodynamics, which is expressed as \( \Delta \mathrm{E} = \mathrm{q} + \mathrm{w} \), the change in internal energy is the sum of heat transferred into or out of the system \( (\mathrm{q}) \) and the work done on or by the system \( (\mathrm{w}) \). In our example of a gas expanding in thermal isolation, understanding how internal energy changes requires looking at both these components, acknowledging that no heat is exchanged due to isolation, leaving work as the driving factor.
Work Done on the System
Work done on or by a system is a crucial aspect in thermodynamics and directly impacts the system's internal energy. Work \( (\mathrm{w}) \) can be considered as energy transfer that causes movement or displacement of a system component, often due to pressure changes. In the context of our problem, the gas performs work as it expands against an external pressure. This work, according to physics conventions, is considered negative work because the system (the gas) is losing energy as it does work on its surroundings.

Therefore, in this scenario, work is calculated based on the pressure-volume work formula, where expanding gas at constant pressure can be represented as \( \mathrm{w} = - P_{\text{ext}} \Delta V \). Since the work is done by the system, the internal energy decreases as energy is expended in moving against the external pressure.
Thermal Isolation
Thermal isolation is a key condition in many thermodynamic problems, where the system does not exchange heat with its surroundings. This is critical because it simplifies the analysis of energy changes. With thermal isolation, the heat transfer \( (\mathrm{q}) \) is zero, implying that any change in the system's internal energy is solely due to the work done.

In our specific example of a gas expanding under thermal isolation, the absence of heat exchange simplifies the First Law of Thermodynamics equation to \( \Delta \mathrm{E} = \mathrm{w} \), where \( \mathrm{q} = 0 \). This assumption allows us to focus entirely on how the work done influences the internal energy, understanding that since work done \( \mathrm{w} \) is negative in this circumstance, it leads to a decrease in internal energy. Such insights are foundational when solving problems that hinge on thermodynamic principles.

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Most popular questions from this chapter

The heat liberated on complete combustion of \(7.8\) g benzene is \(327 \mathrm{~kJ}\). This heat was measured at constant volume and at \(27^{\circ} \mathrm{C}\). Calculate the heat of combustion of benzene at constant pressure ( \(\mathrm{R}\) \(\left.=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) a. \(-3274 \mathrm{~kJ} \mathrm{~mol}^{-1}\) b. \(-1637 \mathrm{~kJ} \mathrm{~mol}^{-1}\) c. \(-3270 \mathrm{~kJ} \mathrm{~mol}^{-1}\) d. \(-3637 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Match the following: List II (Heat of List I \(\quad\) neutralization) A. \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NaOH}\) (p) \(13.7 \mathrm{Kcal}\) B. \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NH}_{4} \mathrm{OH}\) \((\mathrm{q})>13.7 \mathrm{kcal}\) C. \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{4} \mathrm{OH}\) \((\mathrm{r})<13.7 \mathrm{kcal}\) D. \(\mathrm{NaOH}+\mathrm{HF}\) (s) cannot be said

(A): When \(\mathrm{H}_{2} \mathrm{O}\) is added to \(\mathrm{CaO}\), heat is liberated. (R): Reaction between \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O}\) is exothermic.

The \(\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\) for \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) are \(-393.5\), \(-110.5\) and \(-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The standard enthalpy change (in \(\mathrm{kJ}\) ) for the reaction \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is a. \(524.1\) b. \(41.2\) c. \(-262.5\) d. \(-41.2\) [IIT 2000]

(A): Enthalpy changes are positive when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). \(10 \mathrm{H}_{2} \mathrm{O}, \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and salts like \(\mathrm{NaCl}, \mathrm{KCl}\) etc. which do not form hydrates are dissolved in water. But enthalpy changes are negative when anhydrous salts capable of forming hydrates are dissolved in water. (R): The difference in the behaviour is due to large differences in the molecular weights of hydrated and anhydrous salts. The substances with larger molecular weights usually show positive enthalpy change on dissolution.

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